Module Four: Normal distribution and it

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• Module Four Normal distribution and its
  applications to inter-laboratory testing
• When we conduct an inter-laboratory testing, we
  often observe continuous variables,
• e.g., the amount of chloride of a water sample,
  the beta-carotene in a blood sample, the blood
  pressure are continuous variables.
• When we construct a relative frequency histogram,
  it is very likely that the shape of the
  distribution is bell-shaped, that is a few
  possible values are small, a few are large, and
  most of them are around the average.
• Such type of distribution is what we call NORMAL
  distribution.
• Fox example, Blood Pressure, the beta-carotene in
  a blood sample, amount of chloride of a water
  sample mostly follow normal curves.

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A histogram with imposed normal curve for 1900
individuals systolic blood pressure

• The imposed smooth curve looks like a bell-shape.
  If the blood pressure follows a normal curve with
  mean 115 and s.d. 14,
• We use the notation X N(m,s)
• For this case, X N (115,14).
• An immediate question is How can we detect if
  the distribution indeed follows a normal curve.

m115, s 14
Our interest may be to check if the blood
pressure follows a normal distribution, to find
out what proportion of individuals whose blood
pressure is at risk (150 ml or higher), or to
identify extreme cases.
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• When and How do you use Normal Distribution in
  real world situations?
• Normal curve describes the probability of
  occurrences of many real situations.
• Most of statistical techniques, including the
  techniques used for analyzing inter-laboratory
  testing data, assume that the response variable
  approximately follows a normal curve.
• These methods may not be valid if the response
  does not follow a normal distribution. It is,
  therefore, important to learn how to check if a
  response variable follows a normal distribution
  or not. For this reason, we need to learn some
  basic properties of a normal distribution, to
  learn how to compute probabilities and
  percentiles for a normal distribution.
• In this module, we will discuss
• The use of z-table and Minitab to compute
  probabilities and percentiles.
• Techniques of checking if a response variable
  follows a normal distribution.

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• The normal probability distribution provides a
  good model for describing data that have
  mound-shaped frequency distributions.
• The Normal Probability Distribution
• where e 2.718 and p 3.142 m and s (s gt 0
  ) are the parameters that represent the
  population mean and standard deviation.
• We will use the notation X N(m , s). This
  means
• X is distributed as Normal with mean m and
  standard deviation s.
• Some examples of normal random variables are
• X Adult Height , X Scores of s national
  test, X Gas price, X Blood pressure
• NOTE X salary of individuals who are 40 years
  or old before retire does not follow a normal
  curve. It is a skewed to right distribution.

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• Properties of Normal Distribution
• This figure shows three such distributions with
  differing values of m and s .
• Mean determines the center. In this case, m1 lt m2
  lt m3
• Standard deviation measures the variability. In
  this case, s2 lt s1 lt s3
• Large values of s reduce the height of the curve
  and increase the spread.
• Small values of s increase the height of the
  curve and reduce the spread.

s2
s1
s3
m1 m2 m3
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Some properties for X N(m , s)
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• Example
• Every year, universities recruit students using
  their SAT scores. Based on the previous
  information, we know that SAT scores follows a
  normal curve with the mean 1000 and standard
  deviation 180. In the past, CMU admits students
  with SAT 1090 or higher.
• Q1 What is the percent of high school students
  who can receive CMU admission?
• Q2 If CMU decides to higher the SAT admission
  limit to only admit the top 20 of high school
  graduates. What should be the new SAT admission
  limit?
• Q3 A student scored 1200, and claim he is in
  the top 10. Is this a correct claim?

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Tabulated Areas of the Normal Probability
Distributions

• How do you solve the SAT admission problem?
• First, we need to rewrite the problem using the
  notation we are familiar.
• Let call X SAT scores. Then from the given
  information, we know
• X N(1000, 180).
• Q1 asks for P( X gt 1090)
• Q2 asks for a value of X, call it xo, the
  admission limit, so that
• P( X gt xo ) .2
• Q3 asks for comparing P(X gt 1200) with .1
• How do we solve these problems?
• The probability that a continuous random variable
  x assumes a value in the interval from a to b is
  the area under the probability density function
  between the points a and b.

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• One can use computer such as Minitab, or use a
  standardized Z-table.
• The Standard Normal Random Variable
• The standardized normal random variable z, is
  defined as
• z (x - m)/ s , or equivalently, x m zs .
• The standard probability distribution has a mean
  of zero and a standard deviation of 1, that is Z
  N(0,1)
• The area under the standard normal curve between
  mean z 0 and a specified positive value of z,
  say, z0 , is the probability
• Some books use this
• table. Some use other
• type of tables.

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Back to the SAT score problem
X N(1000, 180)
P( Xgt1090)
X, SAT score
1000 1090
Z(x-1000)/180
(1000-1000)/180 0
0.5 (1090-1000)/180
The idea is to transform X N(m,s) to Z(0,1)
using z (x-m)/s P(X gt 1090) P(Z gt
(1090-1000)/180 ) P(Z gt 0.5) Now Z-table can be
applied.
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• Example Find P (0 lt z lt 1.63)
• Solution
• Draw a normal curve, shade the area of interest.
• Rewrite the question in the way that the Z-table
  can be applies. That is in the forms of
• P( 0 lt Z lt zo)
• For this example, it is already in this form, so
  using the Z-table, we obtain P (0 lt z lt 1.63)
  .4484.
• Some additional exercises
• Find P( Z lt 1.96), Find P(-1.24lt Z lt .68), Find
  P( Z gt -1.64)

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• Calculating Probabilities for a General Normal
  Random Variable, X
• 1. Draw a normal curve for X, shade the area of
  interest,
• 2. Transform X to Z.
• - Standardize the interval of interest, write it
  as the equivalent interval in terms of z.
• - The probability of interest is the area that
  you find using the standard normal probability
  distribution.

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Now, Back to the the SAT example, do the
following exercises SAT score, X follows a
normal distribution with mean 1000 and s.d., 180.
That is, X N(1000, 180) Find P(X lt 800) Find
P(750 lt X lt 900) Find P(1180 lt X lt 1360)
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• How about the question of determining the SAT
  admission score for CMU so that the top 20 will
  receive admission from CMU.
• Answer X N(1000, 180). The problem is to find
  the admission score, xo so that
• P(X gt x0) .2
• This is a problem we are looking for a score, not
  a probability. We are reversing the problem
  solving procedure, here.
• Similar technique is applied here
• Draw a normal curve, shade the area of interest.
• Transform from X to Z.
• Rewrite the problem in terms of Z.
• Solve for the standardized value, zo using
  Z-table reversely.
• Transform zo back to xo by xo m s(zo)

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To solve for the admission score xo so that P(X gt
xo) .2 Draw the normal curve, shade the area of
interest, transform to Z. .2 P(X gt xo) P(Z gt
zo) implies P(0 lt Z lt zo) .3 This is a form we
can use Z-table. Looking inside the table, find
the closed probability to .3, which is .2995. By
the Z-table, .2995 P(0 lt Z lt .84). Therefore,
zo .84, which is the standardized admission
limit. So, solving for xo, we have xo m
s(zo) 1000 (180)(.84) 1151.2 The CMU SAT
admission limit will be about 1151.2 (In actual
application for setting up the policy, we can use
1150 as the new admission standard.)
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Hands-on activities Q-aFor the SAT example, X
(1000, 180), suppose a university admits only top
5. Find their admission limit. Q-b Find the 5th
percentile of SAT score. Q-c Find the Q3 SAT
score (75th percentile).
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• Use Minitab to compute cumulative probabilities
  and percentiles for a normal distribution
• Go to Calc, choose Probability Distributions,
  then select Normal.
• In the Dialog box, Density probability f(x),
  Cumulative probability P( X lt a) for any given
  a, Inverse cumulative probability is the 100pth
  percentile, xo , so that P(X lt xo) p. Choose
  the one you are computing.
• Enter Mean and s.d.. By default, it is N(0,1).
• To compute cumulative probability, you need to
  provide a values, which may be created and
  recorded in a column, e.g., C3, or simply to
  provide the constant a.
• To compute inverse cumulative probability, you
  need to provide the cumulative probabilities,
  which must be in (0,1).

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• Methods for detecting the discrepancy of the
  distribution of a response variable from normal
  distribution.
• Consider the example of Blood Pressure data. From
  the histogram and the normal curve imposed onto
  the histogram using Minitab, we can see that the
  blood pressure generally speaking follows a
  normal curve. However, there seems to have a few
  unusually high blood pressures. The question is
  How well the blood pressure follows a normal
  curve?.
• The imposing normal curve helps us to quickly
  identify serious discrepancy from normal.
  However, if the discrepancy is not very serious,
  it is difficult to simply observe the shape of a
  histogram.
• We will discuss three ways for checking the
  normality of a response
• Imposing normal curve onto the histogram,
• Probability plot,
• Numerical methods for testing the degree of
  departure from normal.

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• Imposing a normal curve onto a histogram for the
  blood pressure data of 1900 young adults between
  15-20 years old

The normal curve indicates there are a few large
blood pressure measurements. In fact, the
descriptive statistics shows the highest is 210,
which is much higher than 2 s.d. from the
average. It suggests 210 is very rare. One should
check immediately if there is a typo or not.

• How to construct this plot using Minitab
• Go to Stat, choose Basic Statistics, choose
  Display Descriptive Statistics.
• Enter the variable. Click on the Graphs option,
• In the Graphs option Dialog, you can have a
  variety of choices. One of them is Histogram with
  Normal Curve.

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• 2. Normal Probability Plot It is a
  two-dimensional plot.
• The Y-axis is the estimated cumulative
  probabilities computed by
• The X-axis is the original data in ascending
  order.
• Diagnosis

When the data follow a normal curve, the dotted
points should follow a straight line
When data are skewed-to-right, the plot would
look like
When data are skewed-to-left, the plot would look
like
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• Based on the Normal probability plot, it
  indicates that the systolic blood pressure does
  not follow a normal curve. The pattern also shows
  that the distribution is somewhat
  skewed-to-the-right.
• 3. Test statistic for testing if the blood
  pressure follows a normal curve or not.
• Graphical methods are good to show the pattern
  and gives us pretty clear picture that the data
  do not follow normal. Numerically, there are
  methods that will test such a hypothesis. The
  test statistic is given in the same graph of the
  Normal Probability Plot.
• The Anderson-Darlings Normality Test is
  presented here. The AD-value 11.5, and the
  corresponding p-value is .000
• Note p-value tells us how far the distribution
  of blood pressure is away from normal. The
  smaller the p-value, the less likely the response
  variable follows a normal curve. A common cut-off
  point is 5. In this case, p-value .000, which
  is clear that the distribution of Systolic blood
  pressure does not follow normal.

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• How to construct a Normal Probability Plot and
  carry out the Anderson-Darlings Normality Test?
• Go to Stat, choose Basic Statistics, then select
  Normality Test.
• In the Dialog, enter variable name.
• Reference Probabilities allow us to provide a
  column of cumulative probabilities so that the
  normal probability plot will show the percentiles
  for each given cumulative probability.

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• Note As we have observed that all three methods
  give us similar results. Therefore, the systolic
  blood pressure for 15 to 20 years old young
  adults does not follow a normal distribution from
  the 1909 cases.
• Note Once we find out the distribution is not
  normal, it is critical to take some further
  analysis
• carefully check the data to see if there are any
  typos,
• Examine the data using some descriptive measures
  or other plots to identify extreme cases (Details
  will be discussed in another module).
• Hands-on Activity
• Use the above three methods to check the
  distribution of Diastolic Blood Pressure data.

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• Actions to deal with extreme cases
• For observational studies (such as survey)
• The sample sizes are usually large, and that it
  is often impossible to find out possible causes
  that resulted the extreme data after the data are
  collect. Therefore, it is critical to collect
  background and environmental variables that may
  have potential impact to the results.
• For experimental studies, such as
  inter-laboratory testing
• It is important to look for possible causes that
  resulted the extremes. The study is usually
  conducted under a controlled experimental
  environment. It is more likely to find out causes
  for the extremes, or be able to explain the
  possible causes.
• Deletion of extremes Vs. Making transformation
  to normal
• One must be careful of deleting extremes.
  Especially when we are not able to find any
  causes and the values are reasonable within the
  context of the study.
• This may be an indication that the distribution
  of the response is skewed. For situations such as
  this, an appropriate approach is to transform the
  data to be closer to normal.

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• Method for transforming a variable to normal
• When the data show a skewed distribution,
  statistical methods such as Analysis of Variance
  may not be valid. An approach is to make a
  mathematical transformation of the variable so
  that the transformed variable will be closer to
  normal.
• Some tips for variable transformation
• If variable, Y, is skewed-to-right Then,
  ln(Y), log10(Y), or will be closer of
  normal. (If there are zeros, add each data value
  by .5, first.
• If variable, Y, is skewed-to-left ln(1/Y),
  log10(1/Y),
• or Ya, a gt1 will be closer to normal.

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• An example of Transformation
• The life time of 50 light bulbs are tested by
  letting them on all the time until it burns out.
  The data recorded (in months). Here are the
  histogram and the normal probability test of the
  raw data, the ln transformed data and Square-root
  transformed data

The raw data is skewed-to-right. The Ln
transformation does not work well. The
Square-root transformation works well.
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The normal probability plots and
Anderson-Darlings tests for the life-time data
As the normal probability plots and the Normality
test results indicate, the Sqrt(Y) is
approximately normal. The other two are not.
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Hands-on Activity Analyze the distribution of
variable GR36-Lab-Mean-1 in the TAPPI
inter-laboratory testing study, and determine an
appropriate transformation to make the data
closer to a normal distribution.